SOLUTION: A square has its length increased by 6 feet and its width by 8 feet. If the resulting rectangle has an area of 239.25 square feet what was the perimeter of the original square?

Question 1160800: A square has its length increased by 6 feet and its width by 8 feet. If the resulting rectangle has an area of 239.25 square feet what was the perimeter of the original square?
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Original square,
x and x for length and width
Perimeter: 4x
Area: x^2

Increased square,
x+4 and x+8 for length and width
Perimeter:
Area:

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If the resulting rectangle has an area of 239.25 square feet
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Simplify and solve this for x, before finding the value of 4x.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
s = length of a side of a square.
area of square = S^2.
perimeter of square = 4*S.
L = length of a rectangle.
W = width of a rectangle.
add 6 feet to the length of the square and 8 feet to the width of the square and you get a rectangle that has L = S + 6 and W = S + 8.
you are given that the area of the rectangle is 239.25 square feet.
that means that (S + 6) * (S + 8) = 239.25
simplify this and subtract 239.25 from both sides of the equation and combine like terms and order the terms in descending order of degree and you get:
S^2 + 14*S - 191.25 = 0
factor this quadratic equation to get:
S = -22.5 or 8.5
S can't be negative, so S must be equal to 8.5.
the area of the rectangle would be (8.5 + 6) * (8.5 + 8) = 239.25 square feet, so that checks out ok.
the perimeter of the square would then be 4 * 8.5 = 34 feet.

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